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31 Jan 2022, 05:45
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
46% (02:48) correct
54%
(03:09)
wrong
based on 59
sessions
History
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Three filling pipes R, S and T together can fill an empty tank in 2 hours, S can fill the tank four times faster than T. Initially R alone is opened and after 'x' hours, it is closed and immediately S and T are opened together. The tank is full after another 'y' hours. If the tank was filled in a total of 4 hours, and x is not equal to y, what is the time (in hours) that T alone would take to fill the tank?
(A) 6
(B) 12
(C) 20
(D) 24
(E) 28
Are You Up For the Challenge: 700 Level Questions
(A) 6
(B) 12
(C) 20
(D) 24
(E) 28
Are You Up For the Challenge: 700 Level Questions
Bunuel
First of all, this is a problem solving question so it is bound to have a unique solution.
All three together take 2 hrs so their combines rate is 1/2 tank per hour.
But R is opened alone separately and S and T are opened together but independently of R. The time taken in this case is 4 hrs, twice of time taken before. This happens when the rate of work of 2 individuals/groups is equal. When they work together, time taken is half of time taken when they work independently.
So rate of R must be equal to rate of S and T together.
If rate of T is T, rate of S is 4T (four times rate of T). Then rate of R is 5T (same as combined rate of S and T).
Combined rate of all three = T + 4T + 5T = 10T = 1/2 tank per hour
Rate of T = 1/20th tank per hour
Time taken by T alone is 20 hrs.
Answer (C)
Note that it doesn't matter what x and y are. They could be 1 and 3 respectively or 1.5 and 2.5 or 2.1 and 1.9 etc. Since both the rates are same ('Rate of R alone' = 'Rate of S+T') time taken will be 4 hrs only. Whether we make only R work or S and T work, it has the same effect.
For more on this, check out the work-rate module here using the 3 day trial: https://anaprep.com
General Discussion
31 Jan 2022, 17:14
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Bunuel
Breaking Down the Info:
Let R, S, and T be the rates of the pipes.
R, S and T together can fill an empty tank in 2 hours: \(R + S + T = \frac{1}{2}\)
S can fill the tank four times faster than T: \(S = 5T\), therefore \(S + T = 6T\)
For the next part on x and y, we have \(R*x + (S + T)*y = 1\). Our goal is to find T, so plug out everything else and leave only T and x.
Plug in \(y = 4 - x\) and \(R = \frac{1}{2} - 6T\) to get \((0.5 - 6T)*x + (6T)(4 - x) = 1\)
Simplifying gives \(-12Tx + 24T = 1 - 0.5x\) and \(T = \frac{1 - 0.5x }{ 24 - 12x} = \frac{1}{24}\)
Thus it would take T 24 hours alone.
Answer: D
